Integrand size = 11, antiderivative size = 31 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {1}{12 x^3}-\frac {1}{4 x}-\frac {\left (1+x^2\right )^2 \arctan (x)}{4 x^4} \]
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Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5064, 14} \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {\left (x^2+1\right )^2 \arctan (x)}{4 x^4}-\frac {1}{12 x^3}-\frac {1}{4 x} \]
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Rule 14
Rule 5064
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (1+x^2\right )^2 \arctan (x)}{4 x^4}+\frac {1}{4} \int \frac {1+x^2}{x^4} \, dx \\ & = -\frac {\left (1+x^2\right )^2 \arctan (x)}{4 x^4}+\frac {1}{4} \int \left (\frac {1}{x^4}+\frac {1}{x^2}\right ) \, dx \\ & = -\frac {1}{12 x^3}-\frac {1}{4 x}-\frac {\left (1+x^2\right )^2 \arctan (x)}{4 x^4} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{2 x^2}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-x^2\right )}{12 x^3}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-x^2\right )}{2 x} \]
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Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {\arctan \left (x \right )}{2 x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}-\frac {\arctan \left (x \right )}{4}-\frac {1}{12 x^{3}}-\frac {1}{4 x}\) | \(30\) |
parts | \(-\frac {\arctan \left (x \right )}{2 x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}-\frac {\arctan \left (x \right )}{4}-\frac {1}{12 x^{3}}-\frac {1}{4 x}\) | \(30\) |
parallelrisch | \(-\frac {3 \arctan \left (x \right ) x^{4}+3 x^{3}+6 x^{2} \arctan \left (x \right )+x +3 \arctan \left (x \right )}{12 x^{4}}\) | \(31\) |
meijerg | \(-\frac {1}{12 x^{3}}-\frac {1}{4 x}-\frac {2 \left (-\frac {3 x^{4}}{8}+\frac {3}{8}\right ) \arctan \left (\sqrt {x^{2}}\right )}{3 x^{3} \sqrt {x^{2}}}-\frac {\left (x^{2}+1\right ) \arctan \left (x \right )}{2 x^{2}}\) | \(47\) |
risch | \(\frac {i \left (2 x^{2}+1\right ) \ln \left (i x +1\right )}{8 x^{4}}+\frac {i \left (3 \ln \left (x -i\right ) x^{4}-3 \ln \left (x +i\right ) x^{4}+6 i x^{3}-6 x^{2} \ln \left (-i x +1\right )+2 i x -3 \ln \left (-i x +1\right )\right )}{24 x^{4}}\) | \(80\) |
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {3 \, x^{3} + 3 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right ) + x}{12 \, x^{4}} \]
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Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=- \frac {\operatorname {atan}{\left (x \right )}}{4} - \frac {1}{4 x} - \frac {\operatorname {atan}{\left (x \right )}}{2 x^{2}} - \frac {1}{12 x^{3}} - \frac {\operatorname {atan}{\left (x \right )}}{4 x^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {3 \, x^{2} + 1}{12 \, x^{3}} - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )}{4 \, x^{4}} - \frac {1}{4} \, \arctan \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {3 \, x^{2} + 1}{12 \, x^{3}} - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )}{4 \, x^{4}} - \frac {1}{4} \, \arctan \left (x\right ) \]
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Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {\mathrm {atan}\left (x\right )}{4}-\frac {\frac {x}{12}+\frac {\mathrm {atan}\left (x\right )}{4}+\frac {x^2\,\mathrm {atan}\left (x\right )}{2}+\frac {x^3}{4}}{x^4} \]
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